Question:

A ladder is leaning against a wall. The bottom of the ladder is initially 5 feet away from the wall and is being pulled away at a rate of 2 feet per second. The top of the ladder is 13 feet above the ground and is sliding down the wall at a rate of 1 foot per second. At what rate is the angle between the ladder and the ground changing when the bottom of the ladder is 12 feet away from the wall?

Answer:

Let's denote the distance between the wall and the bottom of the ladder as x (in feet), the height of the ladder as y (in feet), and the angle between the ladder and the ground as θ (in radians).

We are given that the bottom of the ladder is initially 5 feet away from the wall and is being pulled away at a rate of 2 feet per second. This gives us the following information:

dx/dt = 2

We are also given that the top of the ladder is sliding down the wall at a rate of 1 foot per second. This implies that the height of the ladder is changing with respect to time, so:

dy/dt = -1

We are asked to find the rate at which the angle between the ladder and the ground is changing when the bottom of the ladder is 12 feet away from the wall. This implies that we need to find dθ/dt when x = 12.

To solve this problem, we will use the trigonometric relationship between the angle θ, the distance x, and the height y:

tan(θ) = y / x

To differentiate this equation with respect to time, we need to apply the chain rule:

sec^2(θ) * dθ/dt = (dy/dt * x - y * dx/dt) / x^2

Now, let's substitute the given values and solve for dθ/dt:

At x = 12 (when the bottom of the ladder is 12 feet away from the wall): dy/dt = -1 x = 12 dx/dt = 2

Plugging these values into the equation:

sec^2(θ) * dθ/dt = (-1 * 12 - 13 * 2) / 12^2 sec^2(θ) * dθ/dt = (-12 - 26) / 144 sec^2(θ) * dθ/dt = -38 / 144 dθ/dt = (-38 / 144) * (1 / sec^2(θ))

To find sec^2(θ), we can use the Pythagorean Identity:

sec^2(θ) = 1 + tan^2(θ)

We know that tan(θ) = y / x, so we can substitute this into the equation:

sec^2(θ) = 1 + (y / x)^2

At x = 12, we need to find y. Using the Pythagorean Theorem, we can find y:

y^2 = 13^2 - 12^2 y^2 = 169 - 144 y^2 = 25 y = 5

Substituting y = 5 and x = 12 into the equation:

sec^2(θ) = 1 + (5 / 12)^2 sec^2(θ) = 1 + 25 / 144 sec^2(θ) = 169 / 144

Now, substituting sec^2(θ) into the equation for dθ/dt:

dθ/dt = (-38 / 144) * (1 / (169 / 144)) dθ/dt = -38 / 169

Therefore, the rate at which the angle between the ladder and the ground is changing when the bottom of the ladder is 12 feet away from the wall is -38 / 169 radians per second.